1. Calculate pH of the following solutions: a. (H'] = 0.00001M b. (H'] = 2.30 x 10°M c. 0.05M NaOH, 100% ionized d. 0.02M HC,H3O2, a weak acid with Ka = 1.8 x 10 e. (OH'] = 3.5 x 10* M
Ionic Equilibrium
Chemical equilibrium and ionic equilibrium are two major concepts in chemistry. Ionic equilibrium deals with the equilibrium involved in an ionization process while chemical equilibrium deals with the equilibrium during a chemical change. Ionic equilibrium is established between the ions and unionized species in a system. Understanding the concept of ionic equilibrium is very important to answer the questions related to certain chemical reactions in chemistry.
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Arrhenius acid act as a good electrolyte as it dissociates to its respective ions in the aqueous solutions. Keeping it similar to the general acid properties, Arrhenius acid also neutralizes bases and turns litmus paper into red.
Bronsted Lowry Base In Inorganic Chemistry
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Calculate pH of the following solutions:
a. [H+] = 0.00001M
b. [H+] = 2.30 x 10-9 M
c. 0.05M NaOH, 100% ionized
d. 0.02M HC2H3O2, a weak acid with Ka = 1.8 x 10-5
e. [OH-] = 3.5 x 10-8 M
The other picture are the original question and the other is example
im sorry lack of pictures 2 is limit hehe
![Calculate pH of the following solutions:
a. [H'] = 0.00001M
b. (H'] = 2.30 x 10° M
0.05M NaOH, 100% ionized
d. 0.02M HC,H;O2, a weak acid with Ka = 1.8 x 105
C.
e. [OH'] = 3.5 x 10° M
1.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3ee82870-1f5d-43d4-bcac-0c5f3d14ca8b%2F9e7c11a7-1b20-468f-ab02-a7d94ece7923%2Fvomwm9x_processed.png&w=3840&q=75)
![Example 4.1
Calculate the pH of each of the following solutions:
a. [H'] = 2.5 x 1os
b. [OH] = 2.0 x 10%
РОН 3 4.2
d. 0.02M HCI, 100% ionized
e. 0.10M HCN, weak acid with Ka= 4.0 x 1010
C.
Solution:
pH = - log(H'] = - log (2.5 x 10) = - (-4.6) = 4.6
a.
pOH = - log (OH'] = - log (2.0 x 10®) = -(-7.69) = 7.69
pH + pOH = 14
рH 3D14 - рОН
b.
pH = 14 – 7.69 = 6.31
РОН 3 4.2
C.
pH + pОH %3D 14
pH 3D 14 — рОН
= 14 – 4.2
= 9.8
d.
0.02M HCI 100% ionized will have [H*] = 0.02M
pH = - log (H'] = - log (0.02) = - (-1.69) = 1.69](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3ee82870-1f5d-43d4-bcac-0c5f3d14ca8b%2F9e7c11a7-1b20-468f-ab02-a7d94ece7923%2Fp1fpx2j_processed.png&w=3840&q=75)
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